Problem: Simplify and expand the following expression: $ \dfrac{4q - 2}{3q + 2}-\dfrac{q + 1}{q - 10} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3q + 2)(q - 10)$ Multiply the first term by $\dfrac{q - 10}{q - 10}$ $ \begin{align*} \dfrac{4q - 2}{3q + 2} \times \dfrac{q - 10}{q - 10} & = \dfrac{(4q - 2)(q - 10)}{(3q + 2)(q - 10)} \\ & = \dfrac{4q^2 - 42q + 20}{(3q + 2)(q - 10)}\end{align*} $ Multiply the second term by $\dfrac{3q + 2}{3q + 2}$ $ \begin{align*} \dfrac{q + 1}{q - 10} \times \dfrac{3q + 2}{3q + 2} & = \dfrac{(q + 1)(3q + 2)}{(q - 10)(3q + 2)} \\ & = \dfrac{3q^2 + 5q + 2}{(q - 10)(3q + 2)}\end{align*} $ Now we have: $ = \dfrac{4q^2 - 42q + 20}{(3q + 2)(q - 10)} - \dfrac{3q^2 + 5q + 2}{(q - 10)(3q + 2)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{4q^2 - 42q + 20 - (3q^2 + 5q + 2)}{(3q + 2)(q - 10)} $ $ = \dfrac{4q^2 - 42q + 20 - 3q^2 - 5q - 2}{(3q + 2)(q - 10)} $ $ = \dfrac{q^2 - 47q + 18}{(3q + 2)(q - 10)}$ Expand the denominator: $ = \dfrac{q^2 - 47q + 18}{3q^2 - 28q - 20}$